ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If angle DBC = 70 degrees, angle BAC = 30 degrees, find angle BCD. Further, if AB = BC, find angle ECD.
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Here, ∠BCD = ∠BAC
[angles of the same segment]
⇒ ∠BDC = 30° [∵ ∠BAC = 30°]
In ∆BCD;
∠BCD + ∠DBC + ∠BDC = 180°
⇒ ∠BCD + 70° + 30° = 180° [∵ ∠DBC = 70°]
⇒ ∠BCD = 180° - 70° - 30°
⇒ ∠BCD = 80° ---- (1)
If AB = BC
then in ∆ABC --
∠ACB = ∠BAC
[angles opposite to equal sides of a triangle are equal]
⇒ ∠ACB = 30° ----- (2)
Now ∠BCD = ∠ACB + ∠ACD
⇒ 80° = 30° + ∠ACD [using (1) and (2)]
⇒ 80° - 30° = ∠ACD
⇒ 50° = ∠ACD
⇒ ∠ACD = 50°
Hence, ∠ECD = 50°
[angles of the same segment]
⇒ ∠BDC = 30° [∵ ∠BAC = 30°]
In ∆BCD;
∠BCD + ∠DBC + ∠BDC = 180°
⇒ ∠BCD + 70° + 30° = 180° [∵ ∠DBC = 70°]
⇒ ∠BCD = 180° - 70° - 30°
⇒ ∠BCD = 80° ---- (1)
If AB = BC
then in ∆ABC --
∠ACB = ∠BAC
[angles opposite to equal sides of a triangle are equal]
⇒ ∠ACB = 30° ----- (2)
Now ∠BCD = ∠ACB + ∠ACD
⇒ 80° = 30° + ∠ACD [using (1) and (2)]
⇒ 80° - 30° = ∠ACD
⇒ 50° = ∠ACD
⇒ ∠ACD = 50°
Hence, ∠ECD = 50°
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Answered by
107
Solution:-
Since, Triangle ABC & BCD are on the same base.
=> Angle BAC = Angle BDC.
=> BDC = 30°
By Angle Sum Property of Triangle BCD,
=> B + C + D = 180°
=> 70° + C + 30° = 180°
=> C = 180° - 100°
=> Angle C = 80°
Now,
Since, Triangle ABC is an ISOSCELES Triangle.
=> /_ BAC = /_ BCA
=> /_ BCA = 30°
Now,
/_ BCD = /_ BCA + /_ DCB
=> 80° = 30° + /_ DCB
=> /_ DCB = 50°
=> /_ EDC = 50°
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